Tìm đạo hàm của hàm số \(y = \frac{{\left( {2 - {x^2}} \right)\left( {3 - {x^3}} \right)}}{{{{\left( {1 - x} \right)}^2}}}\)
\(\begin{array}{l}
y = \frac{{\left( {2 - {x^2}} \right)\left( {3 - {x^3}} \right)}}{{{{\left( {1 - x} \right)}^2}}} = \frac{{{x^5} - 2{x^3} - 3{x^2} + 6}}{{{{\left( {1 - x} \right)}^2}}}\\
\Rightarrow y\prime = \frac{{\left( {5{x^4} - 6{x^2} - 6x} \right){{\left( {1 - x} \right)}^2} + \left( {{x^5} - 2{x^3} - 3{x^2} + 6} \right).2\left( {1 - x} \right)}}{{{{\left( {1 - x} \right)}^4}}}\\
= \frac{{\left( {1 - x} \right)\left[ {\left( {5{x^4} - 6{x^2} - 6x} \right)\left( {1 - x} \right) + 2{x^5} - 4{x^3} - 6{x^2} + 12} \right]}}{{{{\left( {1 - x} \right)}^4}}}\\
= \frac{{\left( {1 - x} \right)\left( { - 3{x^5} + 5{x^4} + 2{x^3} - 6{x^2} - 6x + 12} \right)}}{{{{\left( {1 - x} \right)}^4}}}\\
= \frac{{ - 3{x^5} + 5{x^4} + 2{x^3} - 6{x^2} - 6x + 12}}{{{{\left( {1 - x} \right)}^3}}}
\end{array}\)
-- Mod Toán 11